hdu 1212 Big Number 数论~大数取余,,还记得秦九昭算法吗~~
发布时间:2020-12-14 02:43:59 所属栏目:大数据 来源:网络整理
导读:Big Number Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5238????Accepted Submission(s): 3638 Problem Description As we know,Big Number is always troublesome. But it's really impor
Big NumberTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5238????Accepted Submission(s): 3638
Problem Description
As we know,Big Number is always troublesome. But it's really important in our ACM. And today,your task is to write a program to calculate A mod B.
To make the problem easier,I promise that B will be smaller than 100000. Is it too hard? No,I work it out in 10 minutes,and my program contains less than 25 lines.
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Input
The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000,and B will be smaller than 100000. Process to the end of file.
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Output
For each test case,you have to ouput the result of A mod B.
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Sample Input
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Sample Output
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比如说1564%m=(((1*10%m+5)*10+6)%m+4)%m) ?;
代码:
#include <stdio.h> #include <string.h> #define MAX 1010 int main() { char num[MAX] ; int m ; while(~scanf("%s%d",num,&m) ) { int len = strlen(num) ; int ans = num[0]-'0' ; for(int i = 1 ; i < len ; ++i) { ans = ans*10%m + num[i]-'0'; } printf("%dn",ans%m) ; } return 0; }? ? ? ? 与君共勉 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |